\newproblem{lay:5_2_24}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.2.24}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Show that if $A$ and $B$ are similar, then $\det\{A\}=\det\{B\}$.
}{
   % Solution
	If $A$ and $B$ are similar, then there exists an invertible matrix $P$ such that
	\begin{center}
		$B=P^{-1}AP$
	\end{center}
	Applying the determinant on both sides we have
	\begin{center}
		$\begin{array}{rcl}\det\{B\}&=&\det\{P^{-1}AP\}=\det\{P^{-1}\}\det\{A\}\det\{P\}\\
		   &=&\frac{1}{\det\{P\}}\det\{A\}\det\{P\}=\det\{A\}\end{array}$
	\end{center}
}
\useproblem{lay:5_2_24}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
